Topic Content:
- Equations with Brackets
Hint: To solve an equation containing brackets eliminate the brackets first, collect like terms on separate sides of the equation and then solve the equation.
Example 1.2.1:
Solve the following equations:
i. 8(5p – 1) + 16 = 8(3p – 9)
ii. 2(2y – 5) – (y + 3) = 3(7 – y)
iii. 4(2x + 5) – 2(x – 3) – 8(2x + 4) = 0
iv. 0 = p + 2.5 – (3.6 – p)
Solution:
i. 8(5p – 1) + 16 = 8(3p – 9)
Expand the brackets
40p – 8 + 16 = 24p – 72
Collect like terms
⇒ 40p – 24p = 8 -16 -72
⇒ 16p = 8 – 88
⇒ 16p = – 80
Divide both sides by 16
⇒ \( \frac {16p}{16} = \frac {-80}{16} \)
∴ p = -5
ii. 2(2y – 5) – (y + 3) = 3(7 – y)
Expand the brackets
4y – 10 – y – 3 = 21 – 3y
Collect like terms
i.e. 4y – y + 3y = 21 + 10 + 3
⇒ 3y + 3y = 34
⇒ 6y = 34
Divide both sides by 6
⇒\( \frac {6y}{6} = \frac {34}{6} \)
y = \( \frac{17}{3} \)
∴ y = \( \scriptsize 5 \normalsize \frac{2}{3} \)
iii. 4(2x + 5) − 2(x – 3) – 8(2x + 4) = 0
Expand the brackets
i.e. 8x + 20 – 2x + 6 – 16x – 24 = 0
Collect like terms
⇒ 8x – 2x – 16x = 24 – 20 – 6
⇒ 8x – 18x = 24 – 26
⇒ -10x = -2
Multiply both sides by -1
i.e. \( \scriptsize -10x \; \times \; -1 \; = \; -2 \; \times \; -1 \)
⇒ 10x = 2
Divide both sides by 10
⇒ \( \frac {10x}{10} = \frac {2}{10} \)
\( \therefore \scriptsize x =\normalsize \frac {1}{5} \)
iv. 0 = p + 2.5 – (3.6 – p)
Expand the brackets
i.e. 0 = p + 2.5 – 3.6 + p
Collect like terms
⇒ -2.5 + 3.6 = p + p
⇒ 1.1 = 2p
Divide both sides by 2
⇒ \( \frac {1.1}{2} = \frac {2p}{2} \)
\(\therefore \scriptsize p = \normalsize \frac {1.1}{2} \scriptsize \: or \: p = \normalsize \frac {11}{20}\)